| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a standard C1 area-between-curves question requiring routine differentiation for the tangent, integration of a polynomial, and finding the area between curve and line using trapezium rule or line equation. All techniques are straightforward applications with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 4x^3 + 6x\) | M1, A1 | M1 for attempt to differentiate |
| At \(x = -1\): \(\frac{dy}{dx} = -4 - 6 = -10\) | M1 | Substituting \(x = -1\) into derivative |
| \(y - 6 = -10(x + 1)\), i.e. \(y = -10x - 4\) | A1 | Correct equation of tangent |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{-1}^{2}(x^4 + 3x^2 + 2)\,dx = \left[\frac{x^5}{5} + x^3 + 2x\right]_{-1}^{2}\) | M1, A1 | M1 for attempt to integrate, A1 correct integration |
| \(= \left(\frac{32}{5} + 8 + 4\right) - \left(-\frac{1}{5} - 1 - 2\right)\) | M1 | Correct substitution of limits |
| \(= \frac{33}{5} + 15 = \frac{33 + 75}{5} = \frac{108}{5}\) | A1, A1 | A1 each bracket correct, A1 final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Area of trapezium (under line \(AB\)): \(\frac{1}{2}(6 + 30)(3) = 54\) | M1 | Finding area under the straight line \(AB\) |
| Shaded area \(= 54 - \frac{108}{5} = \frac{270 - 108}{5} = \frac{162}{5}\) | M1, A1 | Subtracting integral from trapezium area |
# Question 3:
## Part (a)
| $\frac{dy}{dx} = 4x^3 + 6x$ | M1, A1 | M1 for attempt to differentiate |
| At $x = -1$: $\frac{dy}{dx} = -4 - 6 = -10$ | M1 | Substituting $x = -1$ into derivative |
| $y - 6 = -10(x + 1)$, i.e. $y = -10x - 4$ | A1 | Correct equation of tangent |
## Part (b)(i)
| $\int_{-1}^{2}(x^4 + 3x^2 + 2)\,dx = \left[\frac{x^5}{5} + x^3 + 2x\right]_{-1}^{2}$ | M1, A1 | M1 for attempt to integrate, A1 correct integration |
| $= \left(\frac{32}{5} + 8 + 4\right) - \left(-\frac{1}{5} - 1 - 2\right)$ | M1 | Correct substitution of limits |
| $= \frac{33}{5} + 15 = \frac{33 + 75}{5} = \frac{108}{5}$ | A1, A1 | A1 each bracket correct, A1 final answer |
## Part (b)(ii)
| Area of trapezium (under line $AB$): $\frac{1}{2}(6 + 30)(3) = 54$ | M1 | Finding area under the straight line $AB$ |
| Shaded area $= 54 - \frac{108}{5} = \frac{270 - 108}{5} = \frac{162}{5}$ | M1, A1 | Subtracting integral from trapezium area |3 The diagram shows a sketch of a curve and a line.\\
\includegraphics[max width=\textwidth, alt={}, center]{c7f38f7e-75aa-4b41-96fd-f38f968c225c-06_520_588_351_742}
The curve has equation $y = x ^ { 4 } + 3 x ^ { 2 } + 2$. The points $A ( - 1,6 )$ and $B ( 2,30 )$ lie on the curve.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the tangent to the curve at the point $A$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int _ { - 1 } ^ { 2 } \left( x ^ { 4 } + 3 x ^ { 2 } + 2 \right) \mathrm { d } x$.
\item Calculate the area of the shaded region bounded by the curve and the line $A B$.\\[0pt]
[3 marks]
$4 \quad$ A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } + 2 x - 6 y - 40 = 0$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2015 Q3 [12]}}